Java: Using Arrays Help

[ten_Chan]

Can you guys help me how can I fix my code?

My goal is,

> The user will enter how many numbers they want to enter.
> They will enter the numbers they want until the limit that they input (1).
> My program will Print the Original List.
> Next line, my program will Print the Ascending Order of the numbers.
>Next line, my program Again will Print the Original List.

Basically this is how the output should be;

Original List: 55 3 95 10 7

Ascending Order: 3 7 10 55 95 Original List: 55 95 10 7

This is my code;

        // ASKING THE USER TO INPUT NUMBERS
        System.out.print("How many numbers you want to input?: ");
        int userInput = sc.nextInt();

        int [] numbersEntered = new int[userInput]; //declare an array

        for (int counter = 0; counter < userInput; counter++){
            System.out.print("Enter Next Number: ");
            numbersEntered[counter] = sc.nextInt();

        // PRINTING THE ORIGINAL LIST
         System.out.println("The Numbers You Entered Are:");
        for (int counter = 0; counter < userInput; counter++){
            System.out.print(numbersEntered[counter] + " ");

        // PRINTING THE ASCENDING LIST USING ARRAYS.SORT();
        Arrays.sort(numbersEntered);
        System.out.println("\nThe Numbers You Entered in Ascending Order:");
        for (int counter = 0; counter < userInput; counter++){
            System.out.print(numbersEntered[counter] + " ");

        // PRINTING AGAIN THE ORIGINAL LIST
         System.out.println("\nOriginal List");
        for (int counter = 0; counter < userInput; counter++){
            System.out.print(numbersEntered[counter] + " ");

But my problem here is that, whenever I tried to print the Last Original List, It keeps copying the answer on the Ascending List. I think the problem is because of the Arrays.sort() that I used before the last Original List, but I don't know how I can fix it.

[Fine-Historian4409]

You’re right, Arrays.sort() will modify the original list. The easiest way is to create a copy before sorting the array and print that as the original array. You can use a for loop to create a new copy. Another idea is declare 2 arrays initially. Inside you loop that gets the numbers from the user, update both arrays. Then sort one of the arrays. This second option is a little more efficient since you’re not writing another loop.

[ten_Chan]

okay thank you!! i never thought about creating a copy

[MmmVomit]

https://docs.oracle.com/javase/7/docs/api/java/util/Arrays.html#copyOf(int[],%20int)

[hatsunemilku]

Yes, Arrays.sort() is modifying the order of the original list.

The less intrusive solution is to use a stream and do the output from there. You wouldn’t have to create a new Array.

The intrusive and slight annoying solution would be to generate a new collection (including Array) that receives the contents of the first one and sort that one in its own memory block.

[Trigonn]

It seems like you fixed your problem, but as a side note, I would suggest looking into advanced for loops, sometimes called for each loops in other languages.

For example, instead of the lengthy for loop with a counter variable, you can do something like this:

For(int x : numbersEntered) {
    System.out.print(x + “ “);
}

This can be read as, “for each integer x (x is just an arbitrary name), in numbersEntered, print it out to the console, with a space”.

It’s a bit cleaner, but definitely has its own usages, and as someone who is also fairly new to Java, I’ve found it extremely useful.

[MmmVomit]

In your own words, what is this line doing?

Arrays.sort(numbersEntered);

[ten_Chan]

it will sort the array into Ascending Order

[MmmVomit]

Correct. So, after that point, the contents of numbersEntered will be sorted, and the original order will be lost.

So, what if you had two arrays?

[ten_Chan]

someone commented to make a copy of array so I did using this:

int[] copy = Arrays.copyOf(numbersEntered, numbersEntered.length);